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Mrrafil [7]
3 years ago
14

every sixth visitor to an animal shelter gets a free animal calendar every twentieth visitor gets a free animal toy. Which visit

or each day will be the first one to get both the calendar and the animal toy?
Mathematics
1 answer:
Akimi4 [234]3 years ago
6 0
Here we have a case of the least common multipl(lcm) of 6 and 20.
Prime numbers 2,3,5,7,11,13,17,19... (natural numbers greater than 1 that has no positive divisors other than 1 and itself) .
lcm(6,20)= 6 20 | 2
                  3 10 | 3
                  1 10 | 2
                       5 | 5
                       1
2*3*2*5=60 The first one to get both calendar and the animal toy will be 60th.
 
Explenation: First we look for the smallest prime number with wich 6 and 20 can be devided by. That is 2. Next is 3. Since 10 is not divisible by 3, we only copy it. Under the 6 we got 1, wich is our goal. Now we continue to devide 10 by prime numbers till we also get 1. We now multiple all divisors and we get the least common multiple.
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Water is leaking out of an inverted conical tank at a rate of 6800 cubic centimeters per min at the same time that water is bein
Bad White [126]

Answer:

The rate at which water is being pumped into the tank is 13,913.27 cubic centimetre per minute

Step-by-step explanation:

Given that, the height of the conical tank is 13 meters and diameter is 3 meters.

Radius = \frac32 meters

\therefore \frac{radius }{height}=\frac {\frac{3}{2}}{13}

\Rightarrow \frac{radius }{height}=\frac {3}{26}

\Rightarrow radius=\frac {3}{26}\times height   ......(1)

The volume of the cone is   V=\frac13 \pi r^2 h

Now putting r=\frac{3}{26} h

V=\frac13 \pi (\frac3{26}h)^2 h

\Rightarrow V=\frac{3}{676} \pi h^3

Differentiating with respect to t

\frac{dV}{dt}=\frac{3}{676}\pi .3h^2 \frac{dh}{dt}

\Rightarrow \frac{dV}{dt}=\frac{9}{676}\pi h^2 \frac{dh}{dt}

Given that, the water level is rising at a rate of 17 cm per minute when the height 1 meter= 100 cm .i.e \frac{dh}{dt}=17 cm/min

Putting \frac{dh}{dt}=17  

\frac{dV}{dt}=\frac{9}{676}\pi h^2 \times 17

\frac{dV}{dt}|_{h=100}=\frac{9}{676}\pi (100)^2 \times 17

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The volume of water increases 7,113 cubic centimetre per minute while 6,800 cubic centimetre per minute is leaking out.

It means the required rate at which water is being pumped into the tank is

=(7,113+6,800)cubic centimetre per minute

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4 0
3 years ago
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